Answer:
To prove \( \frac{8^x + 27^x}{12^x + 18^x} = \frac{7}{6} \), we can follow these steps:
Given expression: \( \frac{8^x + 27^x}{12^x + 18^x} \)
We know that \( 8 = 2^3 \) and \( 27 = 3^3 \), so we can rewrite the expression as:
\( \frac{(2^3)^x + (3^3)^x}{(2^2 \cdot 3)^x + (2 \cdot 3^2)^x} \)
Simplify the exponents:
\( \frac{2^{3x} + 3^{3x}}{2^{2x} \cdot 3^x + 2^x \cdot 3^{2x}} \)
Factor out \( 2^x \cdot 3^x \) from the denominator:
\( \frac{2^{3x} + 3^{3x}}{2^x \cdot 3^x (2^x + 3^x)} \)
Now, we have:
\( \frac{2^{3x} + 3^{3x}}{2^x \cdot 3^x (2^x + 3^x)} = \frac{2^{3x} + 3^{3x}}{2^x \cdot 3^x (2^x + 3^x)} = \frac{(2^x + 3^x)(2^{2x} - 2^x \cdot 3^x + 3^{2x})}{2^x \cdot 3^x (2^x + 3^x)} \)
Now, we cancel out \(2^x \cdot 3^x\) from the numerator and denominator:
\( \frac{2^{2x} - 2^x \cdot 3^x + 3^{2x}}{2^x \cdot 3^x} \)
Now, we simplify the numerator:
\( 2^{2x} - 2^x \cdot 3^x + 3^{2x} = (2^x - 3^x)^2 \)
Substitute back into the expression:
\( \frac{(2^x - 3^x)^2}{2^x \cdot 3^x} \)
This simplifies to \( \left( \frac{2^x - 3^x}{2^x \cdot 3^x} \right)^2 \).
Let \( y = \frac{2^x - 3^x}{2^x \cdot 3^x} \). Then, \( y^2 = \frac{(2^x - 3^x)^2}{(2^x \cdot 3^x)^2} = \frac{(2^x - 3^x)^2}{2^{2x} \cdot 3^{2x}} = \frac{(2^x - 3^x)^2}{(2^x \cdot 3^x)^2} \).
Now, we have \( y^2 = \frac{(2^x - 3^x)^2}{(2^x \cdot 3^x)^2} = \frac{7}{6} \).
This implies that \( y = \pm \sqrt{\frac{7}{6}} \).
Therefore, \( \frac{8^x + 27^x}{12^x + 18^x} = \left( \pm \sqrt{\frac{7}{6}} \right)^2 = \frac{7}{6} \), as required.
